Definition:
Let $U\subseteq \Bbb R^m$ be an open set. Let $f: U \to \Bbb R^n$ be a function and $T: \Bbb R^m \to \Bbb R^n$ be a linear transformation. We say that $f$ is strongly differentiable at $x_0$, with derivative $T$, if for every $\epsilon>0$ there is $\delta >0$ s.t for any $x_1, x_2$ in $ \delta$-neighbourhood of $x_0$ i.e $\|x_i-x_0\|<\delta$ for $i=1,2$ imply $\|f(x_2)-f(x_1)-T(x_2-x_1)\|<\epsilon\|x_2-x_1\|$.
Show that if $f$ is differentiable in a neighbourhood $U$ of $x_0$ & if $Df$ is continuous as a function of $x$ at $x_0$ then $ f$ is strongly differentiable at $x_0$ .
How I have proceeded is:
$\|f(x_2)-f(x_1)-T(x_2-x_1)\|<\|f(x_2)-f(x_1)\|+ \|T(x_2-x_1)\|<\epsilon\|x_2-x_1\| $ [?? why this is true, I know $\|T(x_2-x_1)\|<\epsilon'\|x_2-x_1\|$ because of definition of norm $T$, but why $\|f(x_2)-f(x_1)\|<\epsilon''\|x_2-x_1\|$]
So can anyone help me saying why $\|f(x_2)-f(x_1)\|<\epsilon''\|x_2-x_1\|$?? Which property will be needed here. Thanks for the help in advance.
